 # YORK UNIVERSITY FACULTY OF HEALTH School of Kinesiology and Health Science

```YORK UNIVERSITY
FACULTY OF HEALTH
School of Kinesiology and Health Science
Kinesiology 2050
3.0
Analysis of Data in Kinesiology
Quiz 2 [Version A]
May 13, 2009
This exam is worth 20% of the total course mark.
This is a closed book exam but non-programmable calculators
may be used.
-
NOTE: The answers to the multiple-choice questions are to
be done on the separate answer sheet provided (Scantron
Sheet). Each multiple choice question is worth one mark
unless noted otherwise.
- Time allowed - 45 minutes.
- Neatly print your name, and lab day and time in the space
provided.
Lab Day:________________
Lab Time: _________________
Lab Instructor: _____________________
MARK=
_________
24
1.
The __________ hypothesis states that the sample mean equals the population mean.
a. Experimental
b. Alternative
c. Directional
d. Null
e. Research
2.
If you decide to reject the null hypothesis then the observed differences between the
experimental and control groups were likely to have:
a. occurred because of the experimental group
b. arisen by chance
c. arisen by treatment effects
d. occurred because of the control group.
3.
The true state of affairs [reality] is that the samples are identical. The decision is to not
reject the null hypothesis. What decision has been made?
a. A correct one
b. Acceptance of the null hypothesis
c. Type I error
d. Type II error
4.
Which of the following samples would most likely have the smallest standard error of
the mean?
a. N = 2
b. N = 10
5.
c. N = 30
d. N = 50
Setting a significance level at .05 means
a. 5 times in 100 our results will be true
b. We will reject the null hypothesis 5 times in 100
c. The probability of rejecting a null hypothesis when it is true is 5 times in 100.
d. The probability of rejecting a null hypothesis when it is true is 95 times in 100.
2 of 14
6.
Find the 95% confidence interval for a sample of 100 with a mean of 65 and a standard
deviation of 30.
a. 59.12 – 70.88
b. 54.26 – 72.74
c. 65.0 – 70.88
d. 63.04 – 66.96
7.
A Canadian railway company claims that its trains block crossings no more that 5
minutes per train on the average. The actual times (minutes) that 10 randomly selected
trains block crossings were:
10.4
9.7
6.5
9.5
8.8
11.2
7.2
10.5
8.2
9.3
2
¯ = 9.130 and s = 2.209.You want to test this claim, at the significance level
…giving X
of 0.05 and assuming that the crossing times are normally distributed. What are the value
of the test statistic and the critical value are, respectively?
a. 5.91 and 2.262
b. 8.79 and 1.833
c. 5.91 and 1.833
d. 8.79 and 2.262
e. 1.87 and 2.262
f. 2.78 and 1.833
g. 2.78 and 1.960
8.
The Gallup Poll [an organization that does political surveys], has decided to increase the
people. The effect of this increase is to:
a. reduce the bias of the estimate.
b. increase the standard error of the estimate.
c. reduce the variability of the estimate.
d. increase the confidence interval width for the parameter.
e. have no effect because the population size is the same.
3 of 14
9.
[2 marks] In some courses students are graded on a “normal/bell” curve. For example,
students within 0.5 standard deviations of the mean receive a “C”; between 0.5 and 1.0
standard deviations above the mean receive a “C+”; between 1.0 and 1.5 standard
deviations above the mean receive a “B”; between 1.5 and 2.0 standard deviations
above the mean receive a “B+”, etc. Assume a normal distribution and that the class
mean was 60 with a standard deviation of 10. What are the bounds for a “B” grade and
a. 65 – 75; 24.17%
b. 70 – 75; 18.37%
c. 70 – 75; 9.19%
d. 65 – 75; 12.08%
e. 70 – 75; 6.68%
10. What is a statistical inference?
a. A decision, estimate, prediction, or generalization about the population based on
information contained in a sample.
b. A statement made about a sample based on the measurements in that sample.
c. A set of data selected from a larger set of data.
d. A decision, estimate, prediction or generalization about sample based on information
contained in a population.
e. A set of data that characterizes some phenomenon.
11. You have measured the systolic blood pressure of a random sample of 25 employees of
a company. A 95% confidence interval for the mean systolic blood pressure for the
employees is computed to be (122 - 138). Which of the following statements gives a
valid interpretation of this interval?
a. About 95% of the sample of employees has a systolic blood pressure between 122 and
138.
b. About 95% of the employees in the company have a systolic blood pressure between 122
and 138.
c. If the sampling procedure were repeated many times, then approximately 95% of the
sample means would be between 122 and 138.
d. The probability that the sample mean falls between 122 and 138 is equal to 0.95.
4 of 14
12. [2 marks] The diameter of ball bearings is known to be normally distributed with
unknown mean and variance. A random sample of size 25 gave a mean 2.5 cm. The
95% confidence interval had a length of ± 2 cm. Then:
a. the sample variance is 4.86.
b. the sample variance is 26.03.
c. the population variance is 4.84.
d. the sample variance is 23.47.
13. [2 marks] A physician wants to compare the blood pressures of six patients before and
after treatment with a drug. The blood pressures are as follows:
Patient
1
2
3
4
5
6
Before Drug After Drug
168
171
171
170
182
180
167
173
174
178
170
172
Mean
SD
172
5.48
174
4.05
The physician wants to use a procedure to test if there is a significant change of the blood
pressure before and after taking the drug at the 0.05 level of significance. The absolute
value of the test statistic and the critical value of the test are, respectively:
a. 1.6151 and 1.956
b. 1.6151 and 2.571
c. 0.7192 and 1.96
d. 0.7192 and 1.812
e. 0.7192 and 2.228
14. A researcher is going to conduct an experiment in order to compare two drugs – a new
drug and an old drug. The researcher would like to see whether there is sufficient
evidence to say that the new drug is better than the old drug. In this problem, the
researcher will commit a Type I error if:
a. she concludes that the drugs are equal in effectiveness when in fact the new drug is better.
b. she concludes that the drugs are equal in effectiveness when in fact the old drug is better.
c. she concludes that the old drug is better when in fact the new drug is better.
d. she concludes that the new drug is better when in fact the drugs are equal in effectiveness.
e. she concludes that the old drug is better when in fact the drugs are equal in effectiveness.
5 of 14
15. If the correlation between body weight and annual income were high and positive, we
could conclude that:
a. High incomes cause people to eat more food.
b. Low incomes cause people to eat less food.
c. High income people tend to spend a greater proportion of their income on food than low
income people, on average.
d. High income people tend to be heavier than low income people, on average.
e. High incomes cause people to gain weight.
16. We are interested in finding the linear relation between the number of vitamin
supplements purchased at one time and the cost per bottle. The following data has been
obtained:
X
Y
No. bottles purchased
Cost per bottle
1
55
3
52
6
46
10
32
15
25
Suppose the regression line is Y = - 2.5X + 60. We compute the average price per bottle of
vitamins if 30 bottles are purchased and observe:
a. Y = -15 dollars;
b. Y = 5 dollars;
c. Y = 25 dollars;
d. Can not be correctly determined with the data provided.
17. A sample of the reading scores of 35 fifth grade students in elementary schools in the
GTA has a mean of 82. The standard deviation of the sample is 15. The 95%
confidence interval [rounded off], for the mean reading scores is ____ and the 99%
confidence interval of the mean reading scores is ____.
a. 77 < µ < 87; 75 < µ < 89
b. 79 < µ < 84; 73 < µ < 91
c. 75 < µ < 87; 73 < µ < 91
d. 73 < µ < 89; 71 < µ < 93
e. 80 < µ < 84; 77 < µ < 87
18. From a pool of 25 applicants, five candidates will be chosen. In how many different
ways can the candidates be chosen?
a. 53,130
b. 298,600
c. 6,375,600
d. 9,765,625
e. 297,200
6 of 14
19. The average time it takes for a person to experience pain relief from aspirin is 25
minutes. A new ingredient is added to help speed up relief. Let μ denote the average
time to obtain pain relief with the new product. An experiment is conducted to verify if
the new product is better. What are the null and alternative hypotheses?
a. H0 : μ = 25 & HA : μ = 25
b. H0 : μ = 25 & HA : μ < 25
c. H0 : μ < 25 & HA : μ = 25
d. H0 : μ < 25 & HA : μ > 25
e. H0 : μ = 25 & HA : μ > 25
20. The following are percentages of fat found in 5 samples of each of two brands of ice
cream:
Brand A 5.7 4.5 6.2 6.3 7.3
Brand B 6.3 5.7 5.9 6.4 5.1
Which of the following procedures is appropriate to test the hypothesis of equal average
fat content in the two brands of ice cream?
a. Paired t-test with 4 d.f.
b. Unpaired t-test with 8 d.f.
c. Paired t-test with 8 d.f.
d. Unpaired t-test with 9 d.f.
e. Repeated measures t-test with 8 d.f.
f. Repeated measures t-test with 9 d.f.
7 of 14
21. A t-test was performed on the data in the question above. A summary of the results is
presented below. [You may assume that the data has been entered correctly.] Based
upon this analysis what can you conclude about the two brands of ice cream?
a. There was no significant difference in the fat content of the two brands of ice cream,
(t=.2343, df = 8, p = NS).
b. There was no significant difference in the fat content of the two brands of ice cream,
(t=.2343, df = 4, p = NS).
c. There was a significant difference in the fat content of the two brands of ice cream,
(t=.2343, df = 4, p < .01). Brand A had higher fat content.
d. There was a significant difference in the fat content of the two brands of ice cream,
(t=.2343, df = 4, p < .01). Brand B had higher fat content.
e. There was a significant difference in the fat content of the two brands of ice cream,
(t=.2343, df = 8, p < .01). Brand A had higher fat content.
f. There was a significant difference in the fat content of the two brands of ice cream,
(t=.2343, df = 8, p < .01). Brand B had higher fat content.
t test
P value
The two-tailed P value is 0.8207.
t = 0.2343 with ?? degrees of freedom.
Summary of Data
Parameter:
Mean:
# of points:
Std deviation:
Std error:
Minimum:
Maximum:
Median:
Lower 95% CI:
Upper 95% CI:
Column A
6.000
5
1.020
0.4561
4.500
7.300
6.200
4.734
7.266
*
*
Column B
5.880
5
0.5215
0.2332
5.100
6.400
5.900
5.233
6.527
*
8 of 14
Formula Sheet
¯ =
X
∑X
location of median =
n
Sample standard deviation:
s=
∑Χ
(n + 1)
2
− Ν(Χ 2 )
2
Ν −1
Standard Error of Mean =
std deviation
N
Percentile Rank =
⎡ ⎛ ( x − LL) ⎞
⎤
⎟⎟( fw) + Σfb ⎥
⎢ ⎜⎜
⎢ ⎝ (i ) ⎠
⎥ 100
⎢
⎥
N
⎢
⎥
⎣
⎦
⎛ ( P) ( N ) − Σfb ⎞
⎟⎟(i )
? Score = LL + ⎜⎜
fw
⎝
⎠
95% CI = X + (1.96) (SEM)
df = N - 1
99% CI = X + (2.58) (SEM)
Z=
Χ− X
SD
SD = Variance
Simple probability p ( A) =
Compound Probability
Dependent Events
Independent Events
Not mutually exclusive
Mutually exclusive
T-score = 10z + 50
# B A has occurred
#A
Conditional Probability p ( B A) =
# O A has occurred
#O
p(A and B) = p(A) · p(B/A)
p(A and B) = p(A) · p(B)
p(A or B) = p(A) + p(B) - p (A and B)
p(A or B) = p(A) + p(B)
Methods of Counting
Permutations
nPr
=
n!
(n − r )!
Combinations
nCr
=
n!
(n − r )!r!
Binomial Probability
r n-r
nCrp q
=
n!
p r q n−r
(n − r )!r!
9 of 14
t=
Χ−μ
sΧ
df = N – 1
Independent Samples
t=
Χ1 − Χ 2
⎛ (n1 − 1)( s1 ) 2 + (n2 − 1)( s2 ) 2 ⎞⎛ 1 1 ⎞
⎟⎟⎜⎜ + ⎟⎟
⎜⎜
n1 + n2 − 2
⎠⎝ n1 n2 ⎠
⎝
df = N1 + N2 - 2
Related Samples
t=
r =
Χ1 − Χ 2
df = N - 1
NΣD 2 − (ΣD) 2
N 2 ( N − 1)
NΣXY − ΣX ΣY
df = N – 2
[ NΣX 2 − (ΣX ) 2 ][ N ΣY 2 − (ΣY ) 2 ]
Y = a + bX
⎡ ( O − E − . 5)2 ⎤
χ = ∑⎢
⎥
E
⎢⎣
⎥⎦
2
O = Observed frequency
n
# of cells
df = k - 1 (k is the number of nominal classes of measurement)
E = Expected Frequency =
⎡ (O − E )2 ⎤
χ = ∑⎢
⎥
⎣ E
⎦
2
Expected freq. =
row total X column total
n
df = (R - 1) (C - 1)
R = number of rows in the contingency table
C = number of columns in the contingency table
ANOVA F =
Treatment var iance
Error var iance
df = (k - 1)/(N - k) k = # of conditions
N= # of subjects.
10 of 14
Table A: Z Scores – Areas Under the Normal Curve
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
0
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
Area between mean and z
0.03
0.04
0.05
0.06
0.0120 0.0160 0.0199 0.0239
0.0517 0.0557 0.0596 0.0636
0.0910 0.0948 0.0987 0.1026
0.1293 0.1331 0.1368 0.1406
0.1664 0.1700 0.1736 0.1772
0.2019 0.2054 0.2088 0.2123
0.2357 0.2389 0.2422 0.2454
0.2673 0.2704 0.2734 0.2764
0.2967 0.2995 0.3023 0.3051
0.3238 0.3264 0.3289 0.3315
0.3485 0.3508 0.3531 0.3554
0.3708 0.3729 0.3749 0.3770
0.3907 0.3925 0.3944 0.3962
0.4082 0.4099 0.4115 0.4131
0.4236 0.4251 0.4265 0.4279
0.4370 0.4382 0.4394 0.4406
0.4484 0.4495 0.4505 0.4515
0.4582 0.4591 0.4599 0.4608
0.4664 0.4671 0.4678 0.4686
0.4732 0.4738 0.4744 0.4750
0.4788 0.4793 0.4798 0.4803
0.4834 0.4838 0.4842 0.4846
0.4871 0.4875 0.4878 0.4881
0.4901 0.4904 0.4906 0.4909
0.4925 0.4927 0.4929 0.4931
0.4943 0.4945 0.4946 0.4948
0.4957 0.4959 0.4960 0.4961
0.4968 0.4969 0.4970 0.4971
0.4977 0.4977 0.4978 0.4979
0.4983 0.4984 0.4984 0.4985
0.4988 0.4988 0.4989 0.4989
11 of 14
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
Table B: Critical values for t
12 of 14
Table E: Critical values for r
13 of 14
Calculations:
14 of 14
``` # Cavi-Lipo Ultrasound Lose Fat and Inches Today! Don’t wait weeks to see results! # Patient Procedure Guide Look Better. Feel Better. Be Better. # comparison of sensory qualities of ice cream from rheological 